Question

Can someone help me with eigenvalues and eigenvectors?

Answer 1

I need help with problem number 6

Answer 2

Use:
\[det(A-\lambda I) = 0\]
to find the eigenvalues. For two by two matrices I think you will get two values of lambra (as a quadratic in lambda will come out).

Answer 3

I basically need help with part C

Answer 4

Lambda=0 and 4

Answer 5

Is that correct?

Answer 6

You have to find a vector X such that \[ AX=\lambda X \]

Answer 7

\[\det(A-I\lambda)=\det\left[\begin{matrix}2-\lambda & 2 \\ 2 & 2-\lambda\end{matrix}\right]=4-2\lambda+\lambda^2-4\]sorry gotta eat!

Answer 8

LOL Go Eat:D

Answer 9

For the corresponding eigenvectors, I think it's a case of finding X in
\[A-\lambda I = X\]
when you know the two values of lambda...

Answer 10

Turing its the opposite

Answer 11

you have I*Iambda-A I'm guessing, it's the same
it is the same

Answer 12

oh sorry
\[(A-\lambda I)X = 0\] (the ero matrix)

Answer 13

zero*

Answer 14

ohhh ya that is the one i have in my textbook

Answer 15

Check out this example : http://en.wikipedia.org/wiki/Eigenvalues_and_eigenvectors#Worked_example

Answer 16

Thanks I figured it out :D

Answer 17

For \( \lambda=0 \) reduced matrix is
\[\left[\begin{matrix}1 & 1 \\ 0 & 0\end{matrix}\right]\]which gives an eigenvector of (1,-1).

For \( \lambda=4 \), the reduced matrix is\[\left[\begin{matrix}-1 & 1 \\ 0 & 0\end{matrix}\right]\]whose eigenvector is (1,1).

Answer 18

Thanks Guys :D I really appreciate ur help

Answer 19

You're welcome! :)